Integral de $$$\frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}}$$$ con respecto a $$$x$$$

La calculadora encontrará la integral/primitiva de $$$\frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}}$$$ con respecto a $$$x$$$, mostrando los pasos.

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Tu entrada

Halla $$$\int \frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}}\, dx$$$.

Solución

Reescribe el integrando:

$${\color{red}{\int{\frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}} d x}}} = {\color{red}{\int{\frac{\cos{\left(x \right)}}{- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}} d x}}}$$

Reescribe el numerador y separa la fracción:

$${\color{red}{\int{\frac{\cos{\left(x \right)}}{- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} - \frac{\left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\alpha \right)}}{\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right) \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(\frac{\cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} - \frac{\left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\alpha \right)}}{\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right) \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{\cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} d x} + \int{\left(- \frac{\left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\alpha \right)}}{\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right) \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d x}\right)}}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=\frac{\cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}}$$$:

$$\int{\left(- \frac{\left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\alpha \right)}}{\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right) \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d x} + {\color{red}{\int{\frac{\cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} d x}}} = \int{\left(- \frac{\left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\alpha \right)}}{\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right) \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d x} + {\color{red}{\frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}}}}$$

Sea $$$u=- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}$$$.

Entonces $$$du=\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right)^{\prime }dx = \left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) dx$$$ (los pasos pueden verse »), y obtenemos que $$$\left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) dx = du$$$.

Por lo tanto,

$$\frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} + {\color{red}{\int{\left(- \frac{\left(- \sin{\left(\alpha \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\alpha \right)}}{\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right) \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d x}}} = \frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} + {\color{red}{\int{\left(- \frac{\sin{\left(\alpha \right)}}{u \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{\sin{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}}$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$\frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} + {\color{red}{\int{\left(- \frac{\sin{\left(\alpha \right)}}{u \left(\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}\right)}\right)d u}}} = \frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} + {\color{red}{\left(- \frac{\sin{\left(\alpha \right)} \int{\frac{1}{u} d u}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}}\right)}}$$

La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} - \frac{\sin{\left(\alpha \right)} {\color{red}{\int{\frac{1}{u} d u}}}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} = \frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} - \frac{\sin{\left(\alpha \right)} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}}$$

Recordemos que $$$u=- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}$$$:

$$\frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)} \sin{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} = \frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} - \frac{\ln{\left(\left|{{\color{red}{\left(- \sin{\left(\alpha \right)} \sin{\left(x \right)} + \cos{\left(\alpha \right)} \cos{\left(x \right)}\right)}}}\right| \right)} \sin{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}}$$

Por lo tanto,

$$\int{\frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}} d x} = \frac{x \cos{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}} - \frac{\ln{\left(\left|{\sin{\left(\alpha \right)} \sin{\left(x \right)} - \cos{\left(\alpha \right)} \cos{\left(x \right)}}\right| \right)} \sin{\left(\alpha \right)}}{\sin^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)}}$$

Simplificar:

$$\int{\frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}} d x} = x \cos{\left(\alpha \right)} - \ln{\left(\left|{\cos{\left(\alpha + x \right)}}\right| \right)} \sin{\left(\alpha \right)}$$

Añade la constante de integración:

$$\int{\frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}} d x} = x \cos{\left(\alpha \right)} - \ln{\left(\left|{\cos{\left(\alpha + x \right)}}\right| \right)} \sin{\left(\alpha \right)}+C$$

Respuesta

$$$\int \frac{\cos{\left(x \right)}}{\cos{\left(\alpha + x \right)}}\, dx = \left(x \cos{\left(\alpha \right)} - \ln\left(\left|{\cos{\left(\alpha + x \right)}}\right|\right) \sin{\left(\alpha \right)}\right) + C$$$A


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